Probability Tree Diagrams | Nikita Brooks- Burrows | - Free Printable

Problem Analysis:

We are given a bag containing:
- 4 purple discs (P),
- 5 white discs (W),
- 2 black discs (B).

A disc is drawn from the bag and not replaced, and then another disc is drawn. We need to complete a tree diagram to determine the probabilities of various outcomes and use it to answer specific questions.

Step 1: Total Number of Discs

The total number of discs in the bag is:
\[
4 + 5 + 2 = 11
\]

Step 2: Constructing the Tree Diagram

We will construct the tree diagram step by step, calculating the probabilities for each branch.

#### First Draw:
- Probability of drawing a Purple (P):
\[
P(\text{P on first draw}) = \frac{4}{11}
\]
- Probability of drawing a White (W):
\[
P(\text{W on first draw}) = \frac{5}{11}
\]
- Probability of drawing a Black (B):
\[
P(\text{B on first draw}) = \frac{2}{11}
\]

#### Second Draw (Conditional Probabilities):
After the first draw, the total number of discs decreases by 1, and the composition of the remaining discs changes depending on the color of the first disc drawn.

##### Case 1: First Draw is Purple (P)
- Remaining discs: 3 Purple, 5 White, 2 Black (Total = 10)
- Probability of drawing a Purple (P) next:
\[
P(\text{P | P}) = \frac{3}{10}
\]
- Probability of drawing a White (W) next:
\[
P(\text{W | P}) = \frac{5}{10} = \frac{1}{2}
\]
- Probability of drawing a Black (B) next:
\[
P(\text{B | P}) = \frac{2}{10} = \frac{1}{5}
\]

##### Case 2: First Draw is White (W)
- Remaining discs: 4 Purple, 4 White, 2 Black (Total = 10)
- Probability of drawing a Purple (P) next:
\[
P(\text{P | W}) = \frac{4}{10} = \frac{2}{5}
\]
- Probability of drawing a White (W) next:
\[
P(\text{W | W}) = \frac{4}{10} = \frac{2}{5}
\]
- Probability of drawing a Black (B) next:
\[
P(\text{B | W}) = \frac{2}{10} = \frac{1}{5}
\]

##### Case 3: First Draw is Black (B)
- Remaining discs: 4 Purple, 5 White, 1 Black (Total = 10)
- Probability of drawing a Purple (P) next:
\[
P(\text{P | B}) = \frac{4}{10} = \frac{2}{5}
\]
- Probability of drawing a White (W) next:
\[
P(\text{W | B}) = \frac{5}{10} = \frac{1}{2}
\]
- Probability of drawing a Black (B) next:
\[
P(\text{B | B}) = \frac{1}{10}
\]

Step 3: Completing the Tree Diagram

Using the probabilities calculated above, we can fill in the tree diagram:

```
P (4/11)
/ | \
/ | \
/ | \
P W B
/|\ /|\ /|\
/ | \ / | \ / | \
P W B P W B P W B
/|\ /|\ /|\ /|\ /|\ /|\ /|\ /|\ /|\
/ | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \
P W B P W B P W B P W B P W B P W B P W B

Probabilities:
- P,P: (4/11) * (3/10) = 12/110 = 6/55
- P,W: (4/11) * (5/10) = 20/110 = 2/11
- P,B: (4/11) * (2/10) = 8/110 = 4/55
- W,P: (5/11) * (4/10) = 20/110 = 2/11
- W,W: (5/11) * (4/10) = 20/110 = 2/11
- W,B: (5/11) * (2/10) = 10/110 = 1/11
- B,P: (2/11) * (4/10) = 8/110 = 4/55
- B,W: (2/11) * (5/10) = 10/110 = 1/11
- B,B: (2/11) * (1/10) = 2/110 = 1/55
```

Step 4: Answering the Questions

#### i. Probability of getting One Purple and One White
We need to consider two scenarios:
1. First draw is Purple, second draw is White: \( P(\text{P, W}) = \frac{4}{11} \times \frac{5}{10} = \frac{20}{110} = \frac{2}{11} \)
2. First draw is White, second draw is Purple: \( P(\text{W, P}) = \frac{5}{11} \times \frac{4}{10} = \frac{20}{110} = \frac{2}{11} \)

Adding these probabilities:
\[
P(\text{One Purple and One White}) = \frac{2}{11} + \frac{2}{11} = \frac{4}{11}
\]

#### ii. Probability of getting One Black and One White
We need to consider two scenarios:
1. First draw is Black, second draw is White: \( P(\text{B, W}) = \frac{2}{11} \times \frac{5}{10} = \frac{10}{110} = \frac{1}{11} \)
2. First draw is White, second draw is Black: \( P(\text{W, B}) = \frac{5}{11} \times \frac{2}{10} = \frac{10}{110} = \frac{1}{11} \)

Adding these probabilities:
\[
P(\text{One Black and One White}) = \frac{1}{11} + \frac{1}{11} = \frac{2}{11}
\]

#### iii. Probability of getting The Same Color
We need to consider the scenarios where both draws result in the same color:
1. Both Purple: \( P(\text{P, P}) = \frac{4}{11} \times \frac{3}{10} = \frac{12}{110} = \frac{6}{55} \)
2. Both White: \( P(\text{W, W}) = \frac{5}{11} \times \frac{4}{10} = \frac{20}{110} = \frac{2}{11} \)
3. Both Black: \( P(\text{B, B}) = \frac{2}{11} \times \frac{1}{10} = \frac{2}{110} = \frac{1}{55} \)

Adding these probabilities:
\[
P(\text{Same Color}) = \frac{6}{55} + \frac{2}{11} + \frac{1}{55} = \frac{6}{55} + \frac{10}{55} + \frac{1}{55} = \frac{17}{55}
\]

Final Answers:

\[
\boxed{\frac{4}{11}, \frac{2}{11}, \frac{17}{55}}
\]